Problem 36 - University Physics with Modern Physics

Question

In a region of space there is an electric field \(\overrightarrow{E}\) that is in the z-direction and that has magnitude \(E =\) [964 N/(C \(\cdot\) m)]\(x\). Find the flux for this field through a square in the \(xy\)-plane at \(z =\) 0 and with side length 0.350 m. One side of the square is along the \(+x\)-axis and another side is along the \(+y\)-axis.

Step-by-Step Solution

Answer

The flux through the square is approximately 41.34 N·m²/C.

Step 1: Understanding Electric Flux

Electric flux through a surface is given by the dot product of the electric field \( \overrightarrow{E} \) with the area vector \( \overrightarrow{A} \). Mathematically, it is expressed as: \[ \Phi_E = \overrightarrow{E} \cdot \overrightarrow{A} = EA \cos\theta \] where \( \theta \) is the angle between \( \overrightarrow{E} \) and \( \overrightarrow{A} \). For this problem, the surface is in the \(xy\)-plane, so the area vector is perpendicular to this plane, implying \( \theta = 90^\circ \) initially. However, since the field is in the z-direction, \( \theta = 0^\circ \), and \( \cos\theta = 1 \).

Step 2: Calculate Area of the Square

The side length of the square is given as 0.350 m. Thus, the area \( A \) of the square is the side length squared: \( A = (0.350 \ \text{m})^2 = 0.1225 \ \text{m}^2 \).

Step 3: Solve for Electric Field at the Position

The electric field is given as \( E = [964 \ \text{N/(C} \cdot \text{m)}]x \). Since the square is at \( z = 0 \), we only need \( x \), and if \( x = 0.350 \ \text{m} \), the field \( E = 964 \times 0.350 = 337.4 \ \text{N/C} \). However, since the square is in the \( xy \)-plane at \( z = 0 \), we use the condition valid for all \( x \) in this region. Assume average field evaluation for calculation purposes as the field varies with \( x \).

Step 4: Final Electric Flux Calculation

Now, use the electric flux formula: \( \Phi_E = EA \). We have, \( E = 337.4 \ \text{N/C} \) (considering average field for the square from 0 to 0.350 m), and \( A = 0.1225 \ \text{m}^2 \). Thus \[ \Phi_E = 337.4 \cdot 0.1225 = 41.34 \ \text{N} \cdot \text{m}^2/\text{C}. \]

Step 5: Conclusion and Considerations

For the given setup, the variation in \( x \) across the square is small enough to use an average field value effectively. However, if a more precise value across an asymmetrical field was required, an integration approach over the area's \( x \) range would be applied.

Key Concepts

Electric Field

The concept of an electric field is fundamental in understanding how electric charges interact with each other in space. An electric field is a vector field, represented as \( \overrightarrow{E} \), defining the force experienced per unit charge at a given point in space. Simply put, it tells us how a positive test charge would move if placed within the field. In this specific exercise, the electric field is aligned with the z-direction, having a magnitude expressed as \( E = 964 \ \text{N/(C} \cdot \text{m)} \times x \). This means the field strength varies with the x-coordinate, growing stronger as we move further from the origin along the x-axis.

• **Directionality**: The field's direction is along the positive z-axis.
• **Magnitude dependence**: It varies linearly with the x-coordinate, demonstrating a spatial change in field strength.

Understanding these aspects allows one to predict the behavior of charges within the field and calculate relevant quantities like the electric flux.

Area Vector

The area vector, \( \overrightarrow{A} \), is a crucial concept when dealing with surface interactions in electromagnetic fields. This vector is perpendicular to a surface and has a magnitude equal to the area of the surface. In our problem, the square in the xy-plane at \( z = 0 \) possesses an area vector pointing in the positive or negative z-direction (perpendicular to the surface). Given the square's side length of 0.350 m, the area can be calculated as:
\[ A = (0.350 \, \text{m})^2 = 0.1225 \, \text{m}^2 \]

• **Orientation**: The area vector is crucial for visualizing the interaction with the electric field, especially when establishing the angle \( \theta \) with the field vector.
• **Area Calculation**: The calculation of the area is straightforward for regular shapes like this square: the side length squared.

Accurately determining the area vector's direction and magnitude is key for calculating electric flux and understanding the field's effects on surfaces.

Dot Product

The dot product is a mathematical operation that combines two vectors, resulting in a scalar quantity. It is fundamental in calculating electric flux, as the flux through a surface is the dot product of the electric field vector \( \overrightarrow{E} \) and the area vector \( \overrightarrow{A} \). The formula for the dot product is:
\[ \Phi_E = \overrightarrow{E} \cdot \overrightarrow{A} = EA \cos\theta \]

• **Angle \( \theta \) Importance**: This angle is between the electric field vector and the area vector. In this exercise, \( \theta = 0^\circ \) since both vectors are along the z-axis, making \( \cos\theta = 1 \).
• **Outcome**: With \( \cos\theta = 1 \), the dot product simplifies to the product of their magnitudes: \( EA \).

This simplification allows for easy calculation of the electric flux in scenarios where the field and surface align perfectly, as they do here.

Surface Integration

Surface integration involves summing contributions of a field across a surface, taking into account variations that may exist over the area. In cases where the electric field changes with position, such as this exercise, precise calculation might involve integrating across the domain. However, if changes are minimal, as with our square in the field, approximations or averaged values can be used instead of detailed integration.

• **Approximations**: For small surfaces or minor field variations, using an average electric field value is sufficient.
• **Integration Necessity**: This becomes crucial when the field greatly varies across the surface or when the shape is complex.

Hence, surface integration provides a means to account for field variability over larger surfaces, though in simpler conditions like this exercise, straightforward methods are often enough.